Problem: Simplify the following expression: $\dfrac{40n^2}{32n}$ You can assume $n \neq 0$.
Solution: $ \dfrac{40n^2}{32n} = \dfrac{40}{32} \cdot \dfrac{n^2}{n} $ To simplify $\frac{40}{32}$ , find the greatest common factor (GCD) of $40$ and $32$ $40 = 2 \cdot 2 \cdot 2 \cdot 5$ $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ $ \mbox{GCD}(40, 32) = 2 \cdot 2 \cdot 2 = 8 $ $ \dfrac{40}{32} \cdot \dfrac{n^2}{n} = \dfrac{8 \cdot 5}{8 \cdot 4} \cdot \dfrac{n^2}{n} $ $\phantom{ \dfrac{40}{32} \cdot \dfrac{2}{1}} = \dfrac{5}{4} \cdot \dfrac{n^2}{n} $ $ \dfrac{n^2}{n} = \dfrac{n \cdot n}{n} = n $ $ \dfrac{5}{4} \cdot n = \dfrac{5n}{4} $